Chapter 8.1, Problem 5E

a.

To determine

To determine the time when the particle is moving to the right , to the left, and stopped.

Answer to Problem 5E

The particle will move right when $0\le t<\frac{\pi }{2}\cup \frac{3\pi }{2}<t\le 2\pi$ .

The particle will move left when $\frac{\pi }{2}<t<\pi \cup \pi <t<2\pi$ .

The particle will stop when $t=\frac{\pi }{2},\pi \text{ and }\frac{3\pi }{2}$ .

Explanation of Solution

Given:

The function $v\left(t\right)$ is the velocity in m/sec of a particle moving along x-axis.

  $v\left(t\right)=5{\sin }^{2}t\cos t,0\le t\le 2\pi$

Calculation:

For the particle to move right, the velocity of particle must be greater than $0$ .

Therefore $v\left(t\right)>0$

  $\begin{array}{l}v\left(t\right)>0\\ 5{\sin }^{2}t\cos t>0\\ {\sin }^{2}t>0\text{ and }\cos t>0\\ t\in \left[0,2\pi \right]\cap \left\{\left[0,\frac{\pi }{2}\right)\cup \left(\frac{\pi }{2},2\pi \right]\right\}\\ \because 0\le t<\frac{\pi }{2}\cup \frac{3\pi }{2}<t\le 2\pi \end{array}$

For the particle to move left, the velocity of particle must be less than $0$ ,

Therefore $v\left(t\right)<0$

  $\begin{array}{l}v\left(t\right)<0\\ 5{\sin }^{2}t\cos t<0\\ {\sin }^{2}t<0\text{ and }\cos t<0\\ \therefore \frac{\pi }{2}<t<\pi \cup \pi <t<2\pi \end{array}$

For the particle to stop , its velocity must be equal to $0$

Therefore, $v\left(t\right)=0$

  $\begin{array}{l}v\left(t\right)=0\\ 5{\sin }^{2}t\cos t=0\\ {\sin }^{2}t=0\text{ or }\cos t=0\\ \therefore t=\frac{\pi }{2},\pi \text{ and }\frac{3\pi }{2}\end{array}$

b.

To determine

To find the particle’s displacement for the time interval $0\le t\le 2\pi$ and final position of particle if $s\left(0\right)=3$ .

Answer to Problem 5E

The displacement of particle is $0$ and the final position of the particle is $3m$ from initial position.

Explanation of Solution

Given:

The function $v\left(t\right)$ is the velocity in m/sec of a particle moving along x-axis.

  $v\left(t\right)=5{\sin }^{2}t\cos t,0\le t\le 2\pi$

Calculation:

The displacement of the particle

  $\begin{array}{l}\frac{ds}{dt}=v\left(t\right)\\ s={\displaystyle \int v\left(t\right)}dt\\ s={\displaystyle \underset{0}{\overset{2\pi }{\int }}5{\sin }^{2}t\cos tdt}\\ \text{let }\sin t=u\\ \text{differentiating both side,}\\ \cos tdt=du\\ s={\displaystyle \underset{0}{\overset{2\pi }{\int }}5u^{2}du}\\ s={\left|\frac{5{\sin }^{3}t}{3}\right|}_0^{2\pi }\\ s=0\end{array}$

Hence, the displacement of particle is $0$ .

Now , displacement = final position − initial position

  $\begin{array}{l}\Delta s=s_f-s_0\\ 0=s_f-s\left(0\right)\\ 0=s_f-3\\ s_f=3\end{array}$

Therefore, the final position of the is at $3m$ from initial position.

c.

To determine

To find the total distance traveled by the particle.

Answer to Problem 5E

The total distance traveled by the particle is $6.67m$ .

Explanation of Solution

Given:

The function $v\left(t\right)$ is the velocity in m/sec of a particle moving along x-axis.

  $v\left(t\right)=5{\sin }^{2}t\cos t,0\le t\le 2\pi$

Calculation:

The total distance traveled by the particle is given by

  $s={\displaystyle \int \left|v\left(t\right)\right|dt}$

Therefore,

  $\begin{array}{l}s={\displaystyle \underset{0}{\overset{2\pi }{\int }}\left|5{\sin }^{2}t\cos t\right|dt}\\ s={\displaystyle \underset{0}{\overset{\pi /2}{\int }}(5{\sin }^{2}t\cos t)}dt+{\displaystyle \underset{\pi /2}{\overset{\pi }{\int }}-(5{\sin }^{2}t\cos t)}dt+{\displaystyle \underset{\pi }{\overset{3\pi /2}{\int }}-(5{\sin }^{2}t\cos t)}dt+{\displaystyle \underset{3\pi /2}{\overset{2\pi }{\int }}(5{\sin }^{2}t\cos t)}dt\\ s={\left|\frac{5{\sin }^{3}t}{3}\right|}_0^{\pi /2}-{\left|\frac{5{\sin }^{3}t}{3}\right|}_{\pi /2}^{\pi }-{\left|\frac{5{\sin }^{3}t}{3}\right|}_{\pi }^{3\pi /2}+{\left|\frac{5{\sin }^{3}t}{3}\right|}_{3\pi /2}^{2\pi }\\ s=\frac{5}{3}\left({\sin }^3\frac{\pi }{2}-{\sin }^{3}0\right)-\frac{5}{3}\left({\sin }^3\pi -{\sin }^3\frac{\pi }{2}\right)-\frac{5}{3}\left({\sin }^3\frac{3\pi }{2}-{\sin }^3\pi \right)+\frac{5}{3}\left({\sin }^{3}2\pi -{\sin }^3\frac{3\pi }{2}\right)\\ s=\frac{5}{3}\left(1-0\right)-\frac{5}{3}\left(0-1\right)-\frac{5}{3}\left(-1-0\right)+\frac{5}{3}\left(0-\left(-1\right)\right)\\ s=\frac{20}{3}=6.67\end{array}$

therefore , the total distance traveled by the particle is $6.67m$ .